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Proof step 4 from proof twols (previous step)

m + n0 + 1 < 2, ∀ m. m < n0 ∧ m + n0 < 2 → n0 = 0 ∨ m = 0 ∧ n0 = 1 ⊦ n0 < m, m = 0 ∧ n0 = 0

User applied rule structural induction

And obtained the following new goals

Proof step 5:
0 + n0 + 1 < 2, ∀ m. m < n0 ∧ m + n0 < 2 → n0 = 0 ∨ m = 0 ∧ n0 = 1 ⊦ n0 < 0, 0 = 0 ∧ n0 = 0

Rule argument

indvar: m

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n;%nat freely generated by 0, +1;;